Problem 22 - University Physics with Modern Physics

Question

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0-kg stone is thrown from the basket with an initial velocity of 15.0 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s. (a) How high is the balloon when the rock is thrown? (b) How high is the balloon when the rock hits the ground? (c) At the instant the rock hits the ground, how far is it from the basket? (d) Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer (i) at rest in the basket and (ii) at rest on the ground.

Step-by-Step Solution

Answer

(a) 122.5 m, (b) 22.5 m, (c) 75 m (d) Horizontal: 15 m/s, Vertical: -69 m/s

Step 1: Identify Known Values

First, identify all the given information from the problem statement: the masses of the balloon, basket, and stone are 124 kg, 22 kg, and 1 kg, respectively. The initial velocity of the stone, as measured by a basket occupant, is 15.0 m/s, and it lands 5.00 seconds post-launch. The balloon's constant descent speed is 20.0 m/s. Gravity is approx. 9.8 m/s².

Step 2: Determine the Balloon's Height When Rock Is Thrown

Calculate how far the balloon descends in 5 seconds. Given its constant speed of 20.0 m/s, this descent is: \( d = v \cdot t = 20.0 \ m/s \times 5.00 \ s = 100 \ m \). The stone takes 5.00 seconds to hit the ground; thus, it falls from 122.5 m: \( y = v_i t + \frac{1}{2}gt^2 \Rightarrow 0 + 0.5(9.8)(25) = 122.5 \ m \).

Step 3: Balloon's Height When Rock Hits Ground

Since the balloon descends at a constant 20.0 m/s, find the height after 5 seconds: Hence, 123 + 0 m initial height minus 100 m descent equals 22.5 m.

Step 4: Compute the Distance From Basket When Stone Hits the Ground

The stone's horizontal velocity is 15.0 m/s relative to the basket. Its horizontal distance traveled is: \( d_x = v_x \times t = 15.0 \ m/s \times 5.00 \ s = 75 \ m \).

Step 5: Velocity Components of Rock at Impact (Basket Perspective)

Velocity components in the basket's view: horizontal stays 15.0 m/s, and vertical at constant velocity of -9.8 \( m/s^2 \times 5 \) determines the stone gains negative 49 m/s.

Step 6: Velocity of Rock at Impact (Ground Perspective)

To find the ground-relative velocity, remember its initial vertical speed is -20 m/s while gravity adds -49 m/s, totalling -69 m/s vertically. Horizontally, velocity is still 15 m/s.

Key Concepts

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause them to move. In this problem, kinematics helps us track the motion of both the balloon and the stone. Understanding kinematics means paying close attention to position, velocity, and acceleration.
In our case:

• We know the constant downward velocity of the balloon is 20.0 m/s.
• The stone is initially thrown with a velocity of 15.0 m/s, perpendicular to the balloon's descent.
• Acceleration due to gravity is 9.8 m/s² downward.

These values allow us to calculate how high the balloon is when the rock is thrown and when it impacts the ground. By utilizing kinematic equations, we find that the balloon loses height as the rock takes 5 seconds to fall.
The balloon descends 100 meters during this time, while the stone, influenced by gravity, descends in a parabolic trajectory until it hits the ground.

Relative Motion

Relative motion examines how different observers perceive the movement of an object. Our problem involves observers both in the basket and on the ground, and they perceive the stone's motion differently.
This exercise is a classic example of relative motion. From the basket, which is itself moving downward, the stone is seen to move horizontally at 15.0 m/s.
The balloon's constant motion adds a vertical perspective for an observer on the ground:

• The basket observer sees the stone thrown sideways, unaffected vertically until gravity takes over.
• To someone on the ground, the stone’s vertical velocity includes both the balloon’s downward motion and the acceleration by gravity.

Relative motion allows us to reconcile these differing perspectives.
We adjust the velocities to account for these observations, essential in finding out the stone's actual path.

Velocity Components

Breaking down the velocity into components simplifies the analysis of projectile motion. Each movement can be considered independently in orthogonal directions: horizontal and vertical. Horizontal velocity remains constant if air resistance is ignored, and vertical velocity changes due to acceleration (like gravity):

• The stone’s horizontal component is 15.0 m/s constantly, relative to the basket.
• The vertical speed changes from 0 (initial throw) to effected by gravity, reflected by the equation \( v_y = v_{0y} + gt \).

In our specific problem:

• In the basket's perspective, there's only a 15.0 m/s horizontal and a changing negative vertical velocity due to gravity.
• From the ground's viewpoint, both the basket’s descent and the effect of gravity need added to the vertical speed.

Understanding these components helps us calculate the stone's distance and direction through the air.

Free Fall Motion

Free fall motion describes objects moving only under the force of gravity. Here, once the stone is thrown, it enters free fall. Gravity acts downward and steadily on any free-falling object, with a constant acceleration of 9.8 m/s². In the exercise:

• As the stone is tossed, it follows a curve, descending over 5 seconds.
• Its vertical motion starts at 0 (measured from the throw) and improves by increasing every second due to gravity.
• Both observers see its fall ending after 5 seconds, a major part of its velocity being attained from gravitational acceleration.

Understanding free-fall caters to conclusively assessing how fast and far objects drop after leaving an initial point, disregarding horizontal movement, and solely considering vertical impact moments.